Ask Question
14 April, 04:43

Steam is contained in a closed rigid container with a volume of 1 m3. initially, the pressure and temperature of the steam are 7 bar and 500°c, respectively. the temperature drops as a result of heat transfer to the surroundings. determine

+3
Answers (1)
  1. 14 April, 04:45
    0
    Question:

    1. Steam is contained in a closed rigid container with a volume of 1 m³. Initially, the pressure and temperature of the steam are 7 bar and 500°C, respectively. The temperature drops as a result of heat transfer to the surroundings. Determine

    a) the temperature at which condensation first occurs, in °C,

    b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.

    c) What is the volume, in m³, occupied by saturated liquid at the final state?

    Answer:

    The answers to the question are;

    a) The temperature at which condensation first occurs is 163.056 °C.

    b) The fraction of the total mass that has condensed when the pressure reaches 0.5 bar is 0.844.

    c) The volume, occupied by saturated liquid at the final state is 1.66*10⁻³ m³.

    Explanation:

    To solve the question, we note that we are required check the steam tables for the properties of steam.

    (a) From P*V = n*R*T

    we find the number of moles as

    n = P₁V₁ / (RT₁)

    Where

    n = Number of moles

    P₁ = Initial pressure = 7 bar = 700000 Pa

    V₁ = Volume = 1 m³

    T₁ = Temperature = 500 C = 773.15 K

    R = Universal gas constant = 8.31451 J / (gmol·K)

    Therefore n = (700000 Pa*1 m³) / (8.31451 J / (gmol·K) * 773.15 K)

    n = 108.89 moles

    Molar mass of H₂O = 18.01528 g/mol

    Therefore initial mass of steam = Number of moles * Molar mass

    = 1961.73 g

    From steam tables, condensation first start to occur at

    151.836 °C + ((179.886 °C - 151.836 °C) / (10 bar - 5 bar)) * (2 bar) = 163.056 °C

    Therefore condensation first start to occur at 163.056 °C.

    b) At 0.5 bar we have

    n = P₂V₂ / (RT₂)

    Where

    V₂ = V₁ = Volume = 1 m³

    P₂ = Pressure = 0.5 bar = 50000 Pa

    T₂ = Saturation temperature at 0.5 Pa = 81.3167 °C = 354.4667 K

    R = Universal gas constant = 8.31451 J / (gmol·K)

    n = (50000 Pa * 1 m³) / (354.4667 K*8.31451 J / (gmol·K)) = 16.965 moles

    Mass of steam left at 0.5 bar = 16.965 moles * 18.01528 g/mol = 305.632 g.

    Mass of condensed steam

    = Initial mass of steam at 7 bar - Final mass of steam at 0.5 bar

    = 1961.73 g - 305.632 g = 1656.095 g

    The fraction of the total mass that has condensed at 0.5 bar is given by

    Mass fraction of condensed steam

    = (Mass of condensed steam) / (Initial mass of steam)

    1656.095 g/1961.73 g = 0.844.

    c) The volume occupied by saturated liquid at the final state is given by

    Volume = Mass/Density = (1656.095 g) / (997 kg/m³)

    = (1.656095 kg) / (997 kg/m³) = 1.66*10⁻³ m³.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Steam is contained in a closed rigid container with a volume of 1 m3. initially, the pressure and temperature of the steam are 7 bar and ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers